Free Worksheet on Combining Like Terms | Solving Equations with Variables on Both Sides and Combining Like Terms Worksheet

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Combining Like Terms with Variables on Both Sides Worksheet

I. Find the sum of the following:
(i) 4a + b – 5c; -a + 3b +3c and 6a – 2b + 8c
(ii) -5x + y + 4z; 7x + 2y + 4z and 3x + 4y – 2z
(iii) 6p + 2q + r; -p + 8q – 9r and 5p – q +3r
(iv) 6a + 4b + 3c; a + 5b – 2c and 4a – b + 6c

Solution:

(i) Given 4a + b – 5c; -a + 3b +3c and 6a – 2b + 8c
First arrange the like terms together,
=4a+(-a)+6a+b+3b+(-2b)+(-5c)+3c+8c
=4a-a+6a+b+3b-2b-5c+11c
=10a-a+4b-2b-5c+11c
=9a+2b+6c
Hence, the sum of like terms 4a + b – 5c, -a + 3b +3c and 6a – 2b + 8c is 9a+2b+6c.

(ii) -5x + y + 4z; 7x + 2y + 4z and 3x + 4y – 2z
First arrange the like terms together,
=-5x+7x+3x+y+2y+4y+4z+4z-2z
=-5x+10x+7y+8z-2z
=5x+7y+6z
Hence, the sum of like terms -5x + y + 4z; 7x + 2y + 4z and 3x + 4y – 2z is 5x+7y+6z.
(iii) 6p + 2q + r; -p + 8q – 9r and 5p – q +3r
By arranging the like terms together we get,
=6p+(-p)+5p+2q+8q-q+r-9r+3r
=11p-p+5p+10q-q+4r-9r
=10p+9q-5r
Therefore, the sum of like terms 6p + 2q + r; -p + 8q – 9r and 5p – q +3r is 10p+9q-5r.
(iv) Given 6a + 4b + 3c; a + 5b – 2c and 4a – b + 6c
By arranging the like terms together we get,
=6a+a+4a+4b+5b+(-b)+3c+(-2c)+6c
=11a+9b-b+9c-2c
=11a+8b+7c
Therefore, the sum of like terms 6a + 4b + 3c; a + 5b – 2c and 4a – b + 6c is 11a+8b+7c.


II. Subtract the following:
(i) Subtract -3xy + 4yz – 14zx from 7xy – 9yz + 6zx
(ii) Subtract -8a2 + 9a – 6 from 2a2 – a + 4
(iii)  subtract a2bc – 8b2ca– 5c2ab from 4a2bc + 3b2ca + c2ab.

Solution:

(i) Given 7xy – 9yz + 6zx take -3xy + 4yz – 14zx
=7xy – 9yz + 6zx-(-3xy+4yz-14zx)
=7xy – 9yz + 6zx + 3xy-4yz + 14zx
Arrange the like terms together,
=7xy+3xy-9yz-4yz+6zx+14zx
=10xy-13yz+20zx
Hence, By subtracting -3xy + 4yz – 14zx from 7xy – 9yz + 6zx we get 10xy-13yz+20zx.
(ii) Given 2a2 – a + 4 take -8a2 + 9a – 6
=2a2 – a + 4 -(-8a2 + 9a – 6)
=2a2 – a + 4+8a2 -9a+6
Arrange the like terms together,
=2a2+8a2-a -9a+4+6
=10a2-10a+10
Therefore, By subtracting -8a2 + 9a – 6 from 2a2 – a + 4 we get 10a2-10a+10.
(iii) Given 4a2bc + 3b2ca + c2ab take a2bc – 8b2ca– 5c2ab
=4a2bc + 3b2ca + c2ab-(a2bc – 8b2ca– 5c2ab)
=4a2bc + 3b2ca + c2ab-a2bc +8b2ca+5c2ab
By arranging the like terms together,
=4a2bc-a2bc+3b2ca+8b2ca +c2ab +5c2ab
=3a2bc+11b2ca+6c2ab
Therefore, By subtracting a2bc – 8b2ca– 5c2ab from 4a2bc + 3b2ca + c2ab we get 3a2bc+11b2ca+6c2ab.


III. What should be added to 2x3 – 6x2y + 2xy2 – 4y3 to get 3x3 + x2y – 2xy2 – 6y3?

Solution:

Let M be the polynomial added to 2x3 – 6x2y + 2xy2 – 4y3.
2x3 – 6x2y + 2xy2 – 4y3 + M=3x3 + x2y – 2xy2 – 6y3
M=3x3 + x2y – 2xy2 – 6y3-(2x3 – 6x2y + 2xy2 – 4y3)
M=3x3 + x2y – 2xy2 – 6y3-2x3 +6x2y -2xy2+4y3
M=x3 +7x2y-4xy2-2y3
Therefore, x3 +7x2y-4xy2-2y3 be added to 2x3 – 6x2y + 2xy2 – 4y3 to get 3x3 + x2y – 2xy2 – 6y3.


IV. Subtract 2m – m3 + 2m2 from the sum of 9 + 5m2 – 2m3+3m and 5m3 – 2m2 + m -9.

Solution:

First, we have to find the sum of 9 + 5m2 – 2m3+3m and 5m3 – 2m2 + m – 9.
By arranging the like terms together, we get
=9-9-2m3+5m3+5m2-2m2+3m+m
=3m3+3m2+4m
Now subtract 2m – m3 + 2m2  from 3m3+3m2+4m.
=3m3+3m2+4m-(2m – m3 + 2m2)
=3m3+3m2+4m-2m+m3 -2m2
=4m3+m2+2m
Therefore, By subtracting 2m – m3 + 2m2 from the sum of 9 + 5m2 – 2m3+3m and 5m3 – 2m2 + m -9 we get 4m3+m2+2m.


V. What is the value of(7x2 −2x+8)−(3x2 +8x−7)?

Solution:

Given, 7x2 −2x+8-(3x2 +8x-7)
=7x2 −2x+8-3x2 -8x+7
=7x2 -3x2 -2x-8x+8+7
=4x2-10x+15
Hence, the value of (7x2 −2x+8)−(3x2 +8x−7) is 4x2-10x+15.


VI.  How much is 2x+y-3z greater than 5x-2y+8z?

Solution:

2x+y-3z=5x-2y+8z+a
a=2x+y-3z-5x+2y-8z
a=-3x+3y-11z
Hence, 2x+y-3z greater than 5x-2y+8z by -3x+3y-11z.


VII. Determine the length of the third side of the triangle whose perimeter is 3x2+15xy+3units and the length of the other two sides is 7xy-2x2 and 3x2+6xy-3.

Solution:

Given,
Two sides of a triangle are 7xy-2x2 and 3x2+6xy-3.
The perimeter of the triangle is 3x2+15xy+3.
To find the length of the third side of the triangle we have to add the other two sides and subtract them from the perimeter.
The sum of the other two sides of the triangle is,
=7xy-2x2 + 3x2+6xy-3
=x2+13xy-3
Subtract x2+13xy-3 from the perimeter 3x2+15xy+3, we get
=3x2+15xy+3-(x2+13xy-3)
=3x2+15xy+3-x2-13xy+3
=2x2+2xy+6
Therefore, the length of the third side of the triangle is 2x2+2xy+6.


VIII. A ribbon of length (7xy-3y+2) units is cut into two parts. If the length of the bigger part is (2xy+y) units, find the length of the smaller part of the ribbon?

Solution:

Given,
Total length of ribbon = (7xy-3y+2) units
The length of the bigger part of the ribbon = (2xy+y) units
To find the length of the smaller part of the ribbon, we need to subtract the length of the larger part from the total length of the ribbon.
=7xy-3y+2-(2xy+y)
=7xy-3y+2-2xy-y
=5xy-4y+2
Therefore, the length of the smaller part of the ribbon is 5xy-4y+2.


IX. From the sum of 3x + 2y – z and 4x – 3y + 2z, subtract the sum of 4x – y + z and x – y – 4z?

Solution:

The sum of 3x + 2y – z and 4x – 3y + 2z is
=3x + 2y – z +4x-3y+2z
=3x+4x+2y-3y-z+2z
=7x-y+z
The sum of 4x – y + z and x – y – 4z is
=4x – y + z + x – y – 4z
=4x+x-y-y+z-4z
=5x-2y-3z
Subtract 5x-2y-3z from 7x-y+z.
=7x-y+z-(5x-2y-3z)
=7x-y+z-5x+2y+3z
=7x-5x-y+2y+z+3z
=2x+y+4z


X. The two sides of a rectangle are x3 – 2y3 + z3 and 3x3 + 5y3 + 5z3. Find its perimeter.

Solution:

Given,
The two sides of a rectangle are x3 – 2y3 + z3 and 3x3 + 5y3 + 5z3
we know that the perimeter of the rectangle is 2(l+w).
The sum of the two sides of the rectangle is
=x3 – 2y3 + z3+3x3 + 5y3 + 5z3
=x3+3x3-2y3+ 5y3+z3+5z3
=4x3+3y3+6z3
The perimeter of the rectangle is
=2(4x3+3y3+6z3)
=8x3+6y3+12z3
Hence, the perimeter of the rectangle is 8x3+6y3+12z3.


 

 

 

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